[R] result of mean(v1, v2, v3) of three real number not the same as sum(v1, v2, v3)/3
Henrik Bengtsson
henr|k@bengt@@on @end|ng |rom gm@||@com
Thu May 12 22:39:20 CEST 2022
There's actually another reason why mean(x) and sum(x)/length(x) may
differ, e.g.
x <- c(rnorm(1e6, sd=.Machine$double.eps), rnorm(1e6, sd=1))
mean(x) - sum(x)/length(x)
#> [1] 1.011781e-18
The mean() function calculates the sample mean using a two pass scan
through the data. The first scan calculates the total sum and divides
by the number of (non-missing) values. In the second scan, this
average is refined by adding the residuals towards the first average.
This way numerical precision of mean(x) is higher than
sum(x)/length(x) when there spread of 'x' is large. It also means
that the processing time of mean(x) is roughly twice that of
sum(x)/length(x).
/Henrik
On Thu, May 12, 2022 at 1:22 PM Ivan Krylov <krylov.r00t using gmail.com> wrote:
>
> Eric Berger and Marc Schwartz and David K Stevens probably said it
> better. I was trying to illustrate the way mean() takes its arguments
> using the match.call function.
>
> The sum() function can take individual numbers or vectors and
> sum all their elements, so sum(c(1, 2, 3)) is the same as sum(1, 2, 3),
> or even sum(c(1, 2), 3): they all do what you mean them to do.
>
> The mean() function is different. It may accept many arguments, but
> only the first of them is the vector of numbers you're interested in:
> mean(c(1, 2, 3)) is the correct way to call it. Unfortunately, when you
> give it more arguments and they aren't what mean() expects them to be
> (the second one should be a number in [0; 0.5] and the third one should
> be TRUE or FALSE, see help(mean) if you're curious), R doesn't warn you
> or raise an error condition.
>
> My use of match.call() was supposed to show that by calling mean(a, b,
> c), I pass the number "b" as the "trim" argument to mean() and the
> number "c" as the "na.rm" argument to mean(), which is not what was
> intended here.
>
> --
> Best regards,
> Ivan
>
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